View Full Version : Question about stellar magnitudes

Brady Yoon

2004-May-11, 05:10 AM

I think I need some help with calculating stellar magnitudes and distances. The major problem is I haven't learned what a logarithm is. I heard somewhere it's the inverse of an exponent; could someone explain? And what are the formulas used to calculate apparent magnitude, absolute magnitude, and distance? I can usually do it, but I always forget since I don't what logarithms are. #-o In the science olympiad state competition, I had to calculate the absolute magnitude of Rigel, knowing its distance of 240 pc and an m of 0.1. I almost got burned on that question; luckily, I already knew it. :(

Help would be most appreciated.

Brady Yoon

Kullat Nunu

2004-May-11, 06:42 AM

With logarithms you can solve exponents.

(here ^ is exponent and _ subscript)

a^x = b <=> x = log_a(b)

For example, if you have to solve a equation 10^x = 100.

10^x = 100

log(10^x) = log(100)

x*log(10) = log(100)

x*1 = log(100)

x = 2.

log here is 10-based logarithm (log_10).

ln is e-based (log_e).

They are the most important logarithms.

You can find more about logarithms for example here (http://www.bbc.co.uk/education/asguru/maths/13pure/05exposandlogs/26logarithms/index.shtml).

Kullat Nunu

2004-May-11, 06:57 AM

You can get apparent and absolute magnitudes as well as distance from the distance modulus equation:

m - M = 5*log(r) - 5

where

m = apparent magnitude

M = absolute magnitude

r = distance in parsecs

Usually when distance need to be determined it is written as

r = 10^((5+m-M)/5)

Googling also helps: http://instruct1.cit.cornell.edu/courses/astro101/lec15.htm

Kullat Nunu

2004-May-11, 07:03 AM

Solution to your problem:

m - M = 5*log(r) - 5

M = m + 5 - 5*log(r)

M = 0.1 + 5 - 5*log(240)

M ~ 5.1 - 11.9

M = -6.8

Sounds right.

milli360

2004-May-11, 02:04 PM

With logarithms you can solve exponents.

(here ^ is exponent and _ subscript)

a^x = b <=> x = log_a(b)

As Kullat Nunu says there, logarithms and exponents are related--they're inverse operations. One "undoes" the other.

If you use your calculater and take the logarithm of 2, it will probably return a value of .30103 or thereabouts. That means the logarithm base is ten, so if you use .30103 as an exponent, 10^.30103, your answer will be 2, as you can check. The other base that is often found on calculators is the "natural" base, which is e, 2.71828

Since star magnitudes are exponential (a star 3 magnitudes brighter will be 2.512^3 brighter), logarithms come into play when you are trying to solve for the magnitude. (The value 2.512 was chosen so that a star 5 magnitudes brighter will be exactly 100 times brighter. It's equal to 100^(1/5), which is the same as 10^(2/5). )

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